Honestly, I can't help you with your application, but I can tell you how the simple calculus works (which I'm assuming is what Alvaro is referring to).

To give an extremely simple introduction, a derivative is the instantaneous rate of change at a specific point on a line. The derivative of a function f(x) (which is notated f'(x)) is found by using the difference quotient:

(This picture's from the Wikipedia article on derivatives; just replace a with x.)

So if f(x) = x2, f'(x) = {the limit as h approaches 0 of} ([x+h]2 - x2)/h. This may seem like it would equal 0; however, if the h variable is left in during expansion, you get:

There are a lot of rules for derivatives that make derivation a lot easier, e.g. the power rule. This states that the derivative of a polynomial function's derivative can be found by taking each term, multiplying it by its power, and then reducing the power by 1. Thus the derivative of x3 = 3x2, and the derivative of 2x2 + 3x = 4x + 3, etc.

Now, for Alvaro's function, (1+e-x)-1, there are a few other rules involved; its derivative (I'm pretty sure) is -(1+e-x)-2-e-x.